Adding pure solution to dilute solution

Question

I have seen the suggested questions, but I'm supposed to solve this by introducing 2 variables and solve them simultaneously:

A cooling system in a particular car contains 7.5L of coolant, of which 33.33333...% is antifreeze

This I calculated to be 2.5 Litres

How much of the total solution must be drained and replaced with pure antifreeze so that the cooling system contains 50% antifreeze.

I could not find 2 variables to start with, and solving them simultaneously looks confusing.

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Could you please explain what you do when you solve it? (All methods I tried ended up with x=0.)

I checked the answer key: 1.875L must replaced with pure antifreeze.

Thnx in advanced.

Answer 1

There is no need of two unknowns, you can work this out with just the amount to replace, let $r$.

Initially, $5\ell$ water, and $2.5\ell$ antifreeze.

After replacement, $(5-\frac23r)\ell$ water and $(2.5-\frac13 r+r)\ell$ coolant (because you remove the $\frac23/\frac13$ mixture).

Then the equation to be solved is

$$5-\frac23 r=2.5+\frac23r,$$ giving

$$r=\frac34\cdot2.5=1.875 \ell$$

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Handling the problem with two variables looks pretty artificial to me. With a little bad faith, let $r$ be the amount of mixture removed, and $a$ the amount of pure antifreeze added. Then

$$\begin{cases}5-\dfrac23r=2.5-\dfrac13r+a,\\r=a.\end{cases}.$$

Answer 2

We want to mix our coolant with pure antifreeze so that there is $7.5L$ concentrated coolant. Let $c$ be the amount of coolant mixture we use, and $a$ the amount of pure antifreeze. In that case, we know two things:

1. We want to have $7.5L$ of liquid at the end. This means that $c+a = 7.5$.
2. We want the new mixture to contain $7.5L/2 = 3.75L$ pure antifreeze. The amount of pure antifreeze in the final mix is equal to the amount of antifreeze we put in, plus a third of the amount of the original coolant we use. That means that $c/3 + a = 3.75$

This gives you two equations you can solve to find your $a$.

Answer 3

Hint:

You can use the two variables:

$y=$ residual coolant at $1/3$ concentration of antifreeze after the substitution.

$x=$ quantity of added pure antifreeze

so that we have $y+x=7.5$ ( the first equation).

Now write that the quantity of antifreeze after the substitution have to be $\frac{7.5}{2}$ Using the fact tha you know how much antifreeze is contained in $y$.